To find the probability distribution for the number of green peppers taken, we consider the possible values of the random variable X, which are 0, 1, 2, and 3.

1. Probability of taking 0 green peppers:

The number of ways to choose 3 peppers from the 8 non-green peppers (3 red + 5 yellow) is \(\binom{8}{3} = 56\).

The total number of ways to choose 3 peppers from all 12 is \(\binom{12}{3} = 220\).

Thus, \(P(X=0) = \frac{56}{220} = \frac{14}{55}\).

2. Probability of taking 1 green pepper:

Choose 1 green pepper from 4 and 2 non-green peppers from 8: \(\binom{4}{1} \times \binom{8}{2} = 4 \times 28 = 112\).

Thus, \(P(X=1) = \frac{112}{220} = \frac{28}{55}\).

3. Probability of taking 2 green peppers:

Choose 2 green peppers from 4 and 1 non-green pepper from 8: \(\binom{4}{2} \times \binom{8}{1} = 6 \times 8 = 48\).

Thus, \(P(X=2) = \frac{48}{220} = \frac{12}{55}\).

4. Probability of taking 3 green peppers:

Choose all 3 peppers from the 4 green peppers: \(\binom{4}{3} = 4\).

Thus, \(P(X=3) = \frac{4}{220} = \frac{1}{55}\).

The probability distribution table is: